Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
2.6 Solving Equations Involving Integers
Recall
A variable is a symbol (usually a letter) that represents a number value that varies.
An equation is a mathematical sentence that equates two mathematical expressions.
If a variable in an equation is replaced by a number and you get a true statement (i.e. the equation balances, left
hand side = right hand side), then that number is called a solution of the equation.
Example 1: Is 2 a solution of 3x  7  1 ?
Solution:
If we replace x with 2 , we get
?
3  2   7 1
?
Simplify each side using the order of operations:
6  7 1
?
1 1
A true statement!
So, x  2 is a solution.
You Try It 1: Which of the following numbers, 8,  8,11, or 11 , is a solution of the equation 2 x  8  14 ?
1
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
Recall
The goal of solving an equation is to isolate the variable on one side with a constant on the other side (i.e. get
the variable to have coefficient of 1).
Emphasis was placed on balancing the equation: whatever operation is done to one side of the equation must
also be done to the other.
In section 1.6, we solved equations like 3  x  15 and 3x  15 . For these equations, we took the following
steps:
Solving an Equation
1) Look at the side of the equation that contains the variable.
2) Decide which operation to use.
 Base decision on the given operation in the equation—how are the numbers
“connected” to the variable?
 Choose the inverse of that operation
3) Perform this inverse operation on both sides of the equation to “get rid” of these
numbers so the variable is isolated.
4) Check your solution.
 Replace the variable(s) of the original equation with the number you got for
your solution.
 Use the order of operations to simplify each side.
 Verify that a true statement occurs.
In this section, we will continue to solve similar equations that make use of our knowledge of integers.
Example 2: Solve and check.
a) 6 x  18
Solution:
a) 6 x  18
b) x  6  18
The variable is on the left hand side.
6 is connected to the variable by multiplication.
Undo multiplication by dividing both sides by 6 .
The solution appears to be 3 .
Check:
Replace x with 3 in the original equation.
Simplify each side.
Did the equation balance? Yes! So x  3 is the solution.
2
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
b) x  6  18
The variable is on the left hand side.
6 is connected to the variable by subtraction.
Undo subtraction by adding 6 to both sides.
The solution appears to be 24 .
Check:
Replace x with 24 in the original equation.
Simplify each side.
Did the equation balance? Yes! So x  24 is the solution.
You Try It 2: Solve and check.
a) x  4  16
b) 4 x  16
c) 0  3  x
d) 0  3x
3
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
Watch out for equations such as  x  7 or 12   x ! The variable is not quite isolated since  x is not the
same as x.  x is understood to be 1x .
Example 3: Solve and check.
a)  x  7
Solution:
a)  x  7
b) 12   x
The variable is on the left hand side.
Write in the understood 1 as the coefficient of x.
1 is connected to the variable by multiplication.
Undo multiplication by dividing both sides by 1 .
The solution appears to be 7 .
Check:
Replace x with 7 in the original equation.
Simplify each side.
Did the equation balance? Yes! So x  7 is the solution.
b) 12   x
The variable is on the right hand side.
Write in the understood 1 as the coefficient of x.
1 is connected to the variable by multiplication.
Undo multiplication by dividing both sides by 1 .
The solution appears to be 12 .
Check:
Replace x with 12 in the original equation.
Simplify each side.
Did the equation balance? Yes! So x  12 is the solution.
You Try It 3: Solve and check.
a) 15   x
b)  x  6
4
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
Some equations involve more than one operation. To solve these equations, you will have to use more than one
inverse operation.
Recall the difference between a variable, a coefficient, a variable term, and a constant term.
In the equation 10 x  9  11 , x is the variable, 10 is the coefficient, 10x is the variable term, and 9 and
11 are constant terms.
Solving an Equation Involving More Than One Operation
1) Look at the side of the equation that contains the variable and notice how the
constant term is “connected” to the variable term.
2) Isolate the variable term by adding (or subtracting) the constant term to (or from)
both sides of the equation.
 If the constant term is positive, subtract it from both sides.
 If the constant term is negative, add it to both sides.
3) Isolate the variable by dividing both sides of the equation by the coefficient.
4) Check your solution.
Notice the steps! Division is always the last step in solving the equation!
Example 4: Solve and check. 10 x  9  11
Solution:
10 x  9  11
The variable term is on the left hand side.
9 is connected to the variable term by subtraction.
Undo subtraction by adding 9 to both sides.
10 x  20
10 is connected to the variable by multiplication.
Undo multiplication by dividing both sides by 10 .
The solution appears to be 2 .
Check:
Replace x with 2 in the original equation.
Simplify each side.
Did the equation balance? Yes! So x  2 is the solution.
5
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
You Try It 4: Solve and check. 5 x  4  16
Example 5: Solve and check. 17  5 x  2
Solution:
17  5 x  2
The variable term is on the right hand side.
2 is connected to the variable term by addition.
Undo addition by subtracting 2 from both sides.
15  5x
5 is connected to the variable by multiplication.
Undo multiplication by dividing both sides by 5 .
The solution appears to be 3 .
Check:
Replace x with 3 in the original equation.
Simplify each side.
Did the equation balance? Yes! So x  3 is the solution.
You Try It 5: Solve and check. 17  6 x  5
6
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
Example 6: Solve and check. 4 x  9  9
Solution:
4 x  9  9
The variable term is on the left hand side.
9 is connected to the variable term by addition.
Undo addition by subtracting 9 from both sides.
4 x  0
5 is connected to the variable by multiplication.
Undo multiplication by dividing both sides by 4 .
The solution appears to be 0 .
Check:
Replace x with 0 in the original equation.
Simplify each side.
Did the equation balance? Yes! So x  0 is the solution.
You Try It 6: Solve and check. 6 x  15  15
Example 7: Solve and check. 14  7  x
Solution:
14  7  x
The variable term is on the right hand side.
7 is positive. (Don’t let the negative coefficient trick you!)
Undo positive 7 by subtracting 7 from both sides.
21   x
Write in the understood 1 as the coefficient of x.
1 is connected to the variable by multiplication.
Undo multiplication by dividing both sides by 1 .
The solution appears to be 21 .
7
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
Replace x with 21 in the original equation.
Simplify each side.
Check:
Did the equation balance? Yes! So x  21 is the solution.
You Try It 7: Solve and check. 18  2  x
Applications
Recall the following when solving application problems.
Five Step Word Problem Method
1) Identify a variable.
Choose a letter to represent the unknown quantity of the word problem.
 Avoid one word descriptions that do not give enough information. For example, x
= money is too vague.
 Remember that variables must always represent quantities. For example,
x = nickels is not a quantity.
Are you referring to how many nickels or the value of the nickels?
x = Dara is not a quantity.
Are you referring to Dara’s age, height, amount of money in her account,
how many dogs she has, etc.?
2) Write an equation.
Look for key phrases that can translate into mathematical symbols.
3) Solve the equation.
Isolate the variable to find the value of the unknown quantity.
4) State your answer in a complete sentence.
Make sure your answer includes any necessary units of measurement (ie, $, feet, …).
5) Check your answer.
Does your answer make sense?
Did you answer what the problem asked for?
8
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
Example 8: Four less than three times a number is negative 19. Find the number.
Solution:
1) define variable: Let n = the number.
2) equation:
Use the words of the problem.
“Four less than three times a number is negative 19”
left side
=
right side
“is” means the equal sign.
left hand side: “less than” means subtraction and “three times” means multiply by 3
right hand side:
3) solve:
4) state answer:
5) check: Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 8: Nine more than twice a number is negative 39. Find the number.
1) define variable:
2) equation:
3) solve:
4) state answer:
5) check: Does the answer makes sense?
Did you answer what the problem is asking for?
9
2015 Carreon
Math 40
Prealgebra
Section 2.6 – Solving Equations Involving Integers
Example 9: A student’s bank account is overdrawn. After making a deposit of $120, he finds his account still
overdrawn by $75. What was his balance before he made the deposit?
Solution:
1) define variable: Let x = the original balance in the student’s account
A positive integer represents a positive balance, while a negative number represents a
debt.
A deposit of $120 is represented by +120. An overdrawn amount is represented by 75
2) equation:
This appears to be a story problem, with a beginning, middle, and end.
beginning
original balance
x
3) solve:
middle
=
deposit
120
end
overdrawn amount

75
x  120  75
4) state answer:
5) check: Does the answer makes sense? Yes!
Did you answer what the problem is asking for? Yes!
You Try It 9: A student’s bank account is overdrawn. After making a deposit of $125, she finds her account
still overdrawn by $10. What was her balance before she made the deposit?
1) define variable:
2) equation:
3) solve:
4) state answer:
5) check: Does the answer makes sense?
Did you answer what the problem is asking for?
10
2015 Carreon